7.4: Double-Angle, Half-Angle, and Reduction Formulas

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Learning Objectives

In this section, you will:

Use double-angle formulas to find exact values.
Use double-angle formulas to verify identities.
Use reduction formulas to simplify an expression.
Use half-angle formulas to find exact values.

Picture of two bicycle ramps, one with a steep slope and one with a gentle slope.

Figure 1 Bicycle ramps for advanced riders have a steeper incline than those designed for novices.

Bicycle ramps made for competition (see Figure 1) must vary in height depending on the skill level of the competitors. For advanced competitors, the angle formed by the ramp and the ground should be θθ such that tanθ=53.tanθ=53. The angle is divided in half for novices. What is the steepness of the ramp for novices? In this section, we will investigate three additional categories of identities that we can use to answer questions such as this one.

Using Double-Angle Formulas to Find Exact Values

In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The double-angle formulas are a special case of the sum formulas, where α=β.α=β. Deriving the double-angle formula for sine begins with the sum formula,

sin(α+β)=sinαcosβ+cosαsinβsin(α+β)=sinαcosβ+cosαsinβ

If we let α=β=θ,α=β=θ, then we have

sin(θ+θ)=sinθcosθ+cosθsinθ sin(2θ)=2sinθcosθsin(θ+θ)=sinθcosθ+cosθsinθ sin(2θ)=2sinθcosθ

Deriving the double-angle for cosine gives us three options. First, starting from the sum formula, cos(α+β)=cosαcosβ−sinαsinβ,cos(α+β)=cosαcosβ−sinαsinβ, and letting α=β=θ,α=β=θ, we have

cos(θ+θ)=cosθcosθ−sinθsinθ cos(2θ)=cos2θ−sin2θcos(θ+θ)=cosθcosθ−sinθsinθ cos(2θ)=cos2θ−sin2θ

Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is:

cos(2θ)=cos2θ−sin2θ =(1−sin2θ)−sin2θ =1−2sin2θcos(2θ)=cos2θ−sin2θ =(1−sin2θ)−sin2θ =1−2sin2θ

The second interpretation is:

cos(2θ)=cos2θ−sin2θ =cos2θ−(1−cos2θ) =2cos2θ−1cos(2θ)=cos2θ−sin2θ =cos2θ−(1−cos2θ) =2cos2θ−1

Similarly, to derive the double-angle formula for tangent, replacing α=β=θα=β=θ in the sum formula gives

tan(α+β)=tanα+tanβ1−tanαtanβtan(θ+θ)=tanθ+tanθ1−tanθtanθtan(2θ)=2tanθ1−tan2θtan(α+β)=tanα+tanβ1−tanαtanβtan(θ+θ)=tanθ+tanθ1−tanθtanθtan(2θ)=2tanθ1−tan2θ

DOUBLE-ANGLE FORMULAS

The double-angle formulas are summarized as follows:

sin(2θ)=2sinθcosθsin(2θ)=2sinθcosθ

cos(2θ)=cos2θ−sin2θ =1−2sin2θ =2cos2θ−1cos(2θ)=cos2θ−sin2θ =1−2sin2θ =2cos2θ−1

tan(2θ)=2tanθ1−tan2θtan(2θ)=2tanθ1−tan2θ

HOW TO

Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value.

Draw a triangle to reflect the given information.
Determine the correct double-angle formula.
Substitute values into the formula based on the triangle.
Simplify.

EXAMPLE 1

Using a Double-Angle Formula to Find the Exact Value Involving Tangent

Given that tanθ=−34tanθ=−34 and θθ is in quadrant II, find the following:

ⓐ sin(2θ)sin(2θ)
ⓑ cos(2θ)cos(2θ)
ⓒ tan(2θ)tan(2θ)

Answer

Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-4,0), and (-4,3). The angle at the origin is theta. The angle formed by the side (-4,3) to (-4,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 5.

TRY IT #1

Given sinα=58,sinα=58, with θθ in quadrant I, find cos(2α).cos(2α).

EXAMPLE 2

Using the Double-Angle Formula for Cosine without Exact Values

Use the double-angle formula for cosine to write cos(6x)cos(6x) in terms of cos(3x).cos(3x).

Answer

Analysis

This example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function.

Using Double-Angle Formulas to Verify Identities

Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.

EXAMPLE 3

Using the Double-Angle Formulas to Establish an Identity

Establish the following identity using double-angle formulas:

1+sin(2θ)=(sinθ+cosθ)21+sin(2θ)=(sinθ+cosθ)2

Answer

Analysis

This process is not complicated, as long as we recall the perfect square formula from algebra:

(a±b)2=a2±2ab+b2(a±b)2=a2±2ab+b2

where a=sinθa=sinθ and b=cosθ.b=cosθ. Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent.

TRY IT #2

Establish the identity: cos4θ−sin4θ=cos(2θ).cos4θ−sin4θ=cos(2θ).

EXAMPLE 4

Verifying a Double-Angle Identity for Tangent

Verify the identity:

tan(2θ)=2cotθ−tanθtan(2θ)=2cotθ−tanθ

Answer

Analysis

Here is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show

2tanθ1−tan2θ=2cotθ−tanθ2tanθ1−tan2θ=2cotθ−tanθ

Let’s work on the right side.

2cotθ−tanθ=21tanθ−tanθ(tanθtanθ) =2tanθ1tanθ(tanθ)−tanθ(tanθ) =2tanθ1−tan2θ2cotθ−tanθ=21tanθ−tanθ(tanθtanθ) =2tanθ1tanθ(tanθ)−tanθ(tanθ) =2tanθ1−tan2θ

When using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier.

TRY IT #3

Verify the identity: cos(2θ)cosθ=cos3θ−cosθsin2θ.cos(2θ)cosθ=cos3θ−cosθsin2θ.

Use Reduction Formulas to Simplify an Expression

The double-angle formulas can be used to derive the reduction formulas, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.

We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let’s begin with cos(2θ)=1−2sin2θ.cos(2θ)=1−2sin2θ. Solve for sin2θ:sin2θ:

cos(2θ)=1−2sin2θ2sin2θ=1−cos(2θ) sin2θ=1−cos(2θ)2cos(2θ)=1−2sin2θ2sin2θ=1−cos(2θ) sin2θ=1−cos(2θ)2

Next, we use the formula cos(2θ)=2cos2θ−1.cos(2θ)=2cos2θ−1. Solve for cos2θ:cos2θ:

cos(2θ)=2cos2θ−11+cos(2θ)=2cos2θ1+cos(2θ)2=cos2θ cos(2θ)=2cos2θ−11+cos(2θ)=2cos2θ1+cos(2θ)2=cos2θ

The last reduction formula is derived by writing tangent in terms of sine and cosine:

tan2θ=sin2θcos2θ =1−cos(2θ)21+cos(2θ)2 =(1−cos(2θ)2)(21+cos(2θ)) =1−cos(2θ)1+cos(2θ)Substitute the reduction formulas.tan2θ=sin2θcos2θ =1−cos(2θ)21+cos(2θ)2Substitute the reduction formulas. =(1−cos(2θ)2)(21+cos(2θ)) =1−cos(2θ)1+cos(2θ)

REDUCTION FORMULAS

The reduction formulas are summarized as follows:

sin2θ=1−cos(2θ)2sin2θ=1−cos(2θ)2

cos2θ=1+cos(2θ)2cos2θ=1+cos(2θ)2

tan2θ=1−cos(2θ)1+cos(2θ)tan2θ=1−cos(2θ)1+cos(2θ)

EXAMPLE 5

Writing an Equivalent Expression Not Containing Powers Greater Than 1

Write an equivalent expression for cos4xcos4x that does not involve any powers of sine or cosine greater than 1.

Answer

Analysis

The solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra.

EXAMPLE 6

Using the Power-Reducing Formulas to Prove an Identity

Use the power-reducing formulas to prove

sin3(2x)=[12sin(2x)][1−cos(4x)]sin3(2x)=[ 12sin(2x) ][ 1−cos(4x) ]

Answer

Analysis

Note that in this example, we substituted

1−cos(4x)21−cos(4x)2

for sin2(2x).sin2(2x). The formula states

sin2θ=1−cos(2θ)2sin2θ=1−cos(2θ)2

We let θ=2x,θ=2x, so 2θ=4x.2θ=4x.

TRY IT #4

Use the power-reducing formulas to prove that 10cos4x=154+5cos(2x)+54cos(4x).10cos4x=154+5cos(2x)+54cos(4x).

Using Half-Angle Formulas to Find Exact Values

The next set of identities is the set of half-angle formulas, which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace θθ with α2,α2, the half-angle formula for sine is found by simplifying the equation and solving for sin(α2).sin(α2). Note that the half-angle formulas are preceded by a ±± sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which α2α2 terminates.

The half-angle formula for sine is derived as follows:

sin2θ=1−cos(2θ)2sin2(α2)=1−cos(2⋅α2)2=1−cosα2sin(α2)=±1−cosα2−−−−−−√ sin2θ=1−cos(2θ)2sin2(α2)=1−cos(2⋅α2)2=1−cosα2sin(α2)=±1−cosα2

To derive the half-angle formula for cosine, we have

cos2θ=1+cos(2θ)2cos2(α2)=1+cos(2⋅α2)2 =1+cosα2 cos(α2)=±1+cosα2−−−−−−√ cos2θ=1+cos(2θ)2cos2(α2)=1+cos(2⋅α2)2 =1+cosα2 cos(α2)=±1+cosα2

For the tangent identity, we have

tan2θ=1−cos(2θ)1+cos(2θ)tan2(α2)=1−cos(2⋅α2)1+cos(2⋅α2) =1−cosα1+cosα tan(α2)=±1−cosα1+cosα−−−−−−√ tan2θ=1−cos(2θ)1+cos(2θ)tan2(α2)=1−cos(2⋅α2)1+cos(2⋅α2) =1−cosα1+cosα tan(α2)=±1−cosα1+cosα

HALF-ANGLE FORMULAS

The half-angle formulas are as follows:

sin(α2)=±1−cosα2−−−−−−−−√sin(α2)=±1−cosα2

cos(α2)=±1+cosα2−−−−−−−−√cos(α2)=±1+cosα2

tan(α2)=±1−cosα1+cosα−−−−−−√=sinα1+cosα=1−cosαsinαtan(α2)=±1−cosα1+cosα=sinα1+cosα=1−cosαsinα

EXAMPLE 7

Using a Half-Angle Formula to Find the Exact Value of a Sine Function

Find sin(15∘)sin(15∘) using a half-angle formula.

Answer

Analysis

Notice that we used only the positive root because sin(15o)sin(15o) is positive.

HOW TO

Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.

Draw a triangle to represent the given information.
Determine the correct half-angle formula.
Substitute values into the formula based on the triangle.
Simplify.

EXAMPLE 8

Finding Exact Values Using Half-Angle Identities

Given that tanα=815tanα=815 and αα lies in quadrant III, find the exact value of the following:

ⓐ sin(α2)sin(α2)
ⓑ cos(α2)cos(α2)
ⓒ tan(α2)tan(α2)

Answer

Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-15,0), and (-15,-8). The angle at the origin is alpha. The angle formed by the side (-15,-8) to (-15,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 17.

TRY IT #5

Given that sinα=−45sinα=−45 and αα lies in quadrant IV, find the exact value of cos(α2).cos(α2).

EXAMPLE 9

Finding the Measurement of a Half Angle

Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of θθ formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If tanθ=53tanθ=53 for higher-level competition, what is the measurement of the angle for novice competition?

Answer: $Image of a right triangle with sides 3, 5, and rad34. Rad 34 is the hypotenuse, and 3 is the base. The angle formed by the hypotenuse and base is theta. The angle between the side of length 3 and side of length 5 is a right angle.$

MEDIA

Access these online resources for additional instruction and practice with double-angle, half-angle, and reduction formulas.

7.3 Section Exercises

Verbal

Explain how to determine the reduction identities from the double-angle identity cos(2x)=cos2x−sin2x.cos(2x)=cos2x−sin2x.

Explain how to determine the double-angle formula for tan(2x)tan(2x) using the double-angle formulas for cos(2x)cos(2x) and sin(2x).sin(2x).

We can determine the half-angle formula for tan(x2)=1−cosx√1+cosx√tan(x2)=1−cosx1+cosx by dividing the formula for sin(x2)sin(x2) by cos(x2).cos(x2). Explain how to determine two formulas for tan(x2)tan(x2) that do not involve any square roots.

For the half-angle formula given in the previous exercise for tan(x2),tan(x2), explain why dividing by 0 is not a concern. (Hint: examine the values of cosxcosx necessary for the denominator to be 0.)

Algebraic

For the following exercises, find the exact values of a) sin(2x),sin(2x), b) cos(2x),cos(2x), and c) tan(2x)tan(2x) without solving for x.x.

If sinx=18,sinx=18, and xx is in quadrant I.

If cosx=23,cosx=23, and xx is in quadrant I.

If cosx=−12,cosx=−12, and xx is in quadrant III.

If tanx=−8,tanx=−8, and xx is in quadrant IV.

For the following exercises, find the values of the six trigonometric functions if the conditions provided hold.

cos(2θ)=35cos(2θ)=35 and 90∘≤θ≤180∘90∘≤θ≤180∘

10.

cos(2θ)=12√cos(2θ)=12 and 180∘≤θ≤270∘180∘≤θ≤270∘

For the following exercises, simplify to one trigonometric expression.

11.

2sin(π4)2cos(π4)2sin(π4)2cos(π4)

12.

4sin(π8)cos(π8)4sin(π8)cos(π8)

For the following exercises, find the exact value using half-angle formulas.

13.

sin(π8)sin(π8)

14.

cos(−11π12)cos(−11π12)

15.

sin(11π12)sin(11π12)

16.

cos(7π8)cos(7π8)

17.

tan(5π12)tan(5π12)

18.

tan(−3π12)tan(−3π12)

19.

tan(−3π8)tan(−3π8)

For the following exercises, find the exact values of a) sin(x2),sin(x2), b) cos(x2),cos(x2), and c) tan(x2)tan(x2) without solving for x,x, when 0∘≤x≤360∘0∘≤x≤360∘

20.

If tanx=−43,tanx=−43, and xx is in quadrant IV.

21.

If sinx=−1213,sinx=−1213, and xx is in quadrant III.

22.

If cscx=7,cscx=7, and xx is in quadrant II.

23.

If secx=−4,secx=−4, and xx is in quadrant II.

For the following exercises, use Figure 5 to find the requested half and double angles.

Image of a right triangle. The base is length 12, and the height is length 5. The angle between the base and the height is 90 degrees, the angle between the base and the hypotenuse is theta, and the angle between the height and the hypotenuse is alpha degrees.

Figure 5

24.

Find sin(2θ),cos(2θ),sin(2θ),cos(2θ), and tan(2θ).tan(2θ).

25.

Find sin(2α),cos(2α),sin(2α),cos(2α), and tan(2α).tan(2α).

26.

Find sin(θ2),cos(θ2),sin(θ2),cos(θ2), and tan(θ2).tan(θ2).

27.

Find sin(α2),cos(α2),sin(α2),cos(α2), and tan(α2).tan(α2).

For the following exercises, simplify each expression. Do not evaluate.

28.

cos2(28∘)−sin2(28∘)cos2(28∘)−sin2(28∘)

29.

2cos2(37∘)−12cos2(37∘)−1

30.

1−2sin2(17∘)1−2sin2(17∘)

31.

cos2(9x)−sin2(9x)cos2(9x)−sin2(9x)

32.

4sin(8x)cos(8x)4sin(8x)cos(8x)

33.

6sin(5x)cos(5x)6sin(5x)cos(5x)

For the following exercises, prove the identity given.

34.

(sint−cost)2=1−sin(2t)(sint−cost)2=1−sin(2t)

35.

sin(2x)=−2sin(−x)cos(−x)sin(2x)=−2sin(−x)cos(−x)

36.

cotx−tanx=2cot(2x)cotx−tanx=2cot(2x)

37.

1+cos(2θ)sin(2θ)tan2θ=tanθ1+cos(2θ)sin(2θ)tan2θ=tanθ

For the following exercises, rewrite the expression with an exponent no higher than 1.

38.

cos2(5x)cos2(5x)

39.

cos2(6x)cos2(6x)

40.

sin4(8x)sin4(8x)

41.

sin4(3x)sin4(3x)

42.

cos2xsin4xcos2xsin4x

43.

cos4xsin2xcos4xsin2x

44.

tan2xsin2xtan2xsin2x

Technology

For the following exercises, reduce the equations to powers of one, and then check the answer graphically.

45.

tan4xtan4x

46.

sin2(2x)sin2(2x)

47.

sin2xcos2xsin2xcos2x

48.

tan2xsinxtan2xsinx

49.

tan4xcos2xtan4xcos2x

50.

cos2xsin(2x)cos2xsin(2x)

51.

cos2(2x)sinxcos2(2x)sinx

52.

tan2(x2)sinxtan2(x2)sinx

For the following exercises, algebraically find an equivalent function, only in terms of sinxsinx and/or cosx,cosx, and then check the answer by graphing both equations.

53.

sin(4x)sin(4x)

54.

cos(4x)cos(4x)

Extensions

For the following exercises, prove the identities.

55.

sin(2x)=2tanx1+tan2xsin(2x)=2tanx1+tan2x

56.

cos(2α)=1−tan2α1+tan2αcos(2α)=1−tan2α1+tan2α

57.

tan(2x)=2sinxcosx2cos2x−1tan(2x)=2sinxcosx2cos2x−1

58.

(sin2x−1)2=cos(2x)+sin4x(sin2x−1)2=cos(2x)+sin4x

59.

sin(3x)=3sinxcos2x−sin3xsin(3x)=3sinxcos2x−sin3x

60.

cos(3x)=cos3x−3sin2xcosxcos(3x)=cos3x−3sin2xcosx

61.

1+cos(2t)sin(2t)−cost=2cost2sint−11+cos(2t)sin(2t)−cost=2cost2sint−1

62.

sin(16x)=16sinxcosxcos(2x)cos(4x)cos(8x)sin(16x)=16sinxcosxcos(2x)cos(4x)cos(8x)

63.

cos(16x)=(cos2(4x)−sin2(4x)−sin(8x))(cos2(4x)−sin2(4x)+sin(8x))